NOTES ON RINGS, MATH 369.101 Kernels of ring homomorphisms and Ideals Recall the definition of a ring homomorphism. Some new examples: (1) Complex conjugation: z = a + bi 7! z = a − bi (where i2 = −1). This gives a ring homomorphism C ! C, since we can check that 1 = 1, z1 + z2 = z1 + z2 and z1z2 = z1 z2. To check the last one, let z1 = a + bi; z2 = c + di: z1z2 = (a + bi)(c + di) = ac − bd + (ad + bc)i but z1 z2 = (a − bi)(c − di) = ac − bd − (ad + bc)i = z1z2: (2) Evaluation: For any ring R, choose r 2 R. Then evaluation at x = r gives a ring homomorphism φr : R[x] ! R defined by φr(f(x)) = n f(r). In other words, if f(x) = a0 + a1x + ::: + anx is a polynomial in R[x], then n φr(f(x)) = f(r) = a0 + a1r + ::: + anr : It is the case that f(a) is in R, since the coefficients of f were in R, a 2 R, and all the operations on polynomials are the addition and multiplication operations used in the ring. Think about how evaluation at x = r is defined on a sum f(x) + g(x) and product f(x)g(x) of polynomials, and why this means that φr preserves addition and multiplication. So φr is a homomorphism, even when R is not a commutative ring. As an example of the evaluation homomorphism, think of when 2 R = Z and we choose some integer n 2 Z. Then φn(1 + x + 2x ) = 2 1 + n + 2n . For which f(x) does φn(f(x)) = 0? Can you describe this set of f(x) as \all multiples of some particular polynomial"? P.S. As another nice example of the evaluation homomorphism, one could think of evaluation at a matrix of a polynomial in R[x] where R = Mn(R). The fact that this is a homomorphism provides the essential details for why the Cayley-Hamilton theorem (from linear algebra) is true. Proposition 1. Composition of two ring homomorphisms is a ring homo- morphism. Date: Oct. 12 { Oct. 19. 1 2 NOTES ON RINGS, MATH 369.101 Proof. Let φ : R ! S and : S ! T be two ring homomorphisms. We need to show that ◦ φ (defined by (φ(r))) is a ring homomorphism. First, we check that 1 is sent to 1: (φ(1)) = (1) = 1, the first equality because φ is a ring homomorphism, the second equality because is a ring homomorphism. Second, choose r1; r2 2 R. We check that the composition preserves addition: (φ(r1 +r2)) = (φ(r1)+φ(r2)) = (φ(r1))+ (φ(r2)). Again, the first equality holds because φ is a ring homomorphism, the second equality because is a ring homomorphism. The reason that multiplication is preserved is similar: (φ(r1 · r2)) = (φ(r1) · φ(r2)) = (φ(r1)) · (φ(r2)). So we can compose two homomorphisms and still have a homomorphism. Exercise. Think about the rings Z9 and Z3 ⊕ Z3 and a ring homomor- phism Z9 ! Z3 ⊕ Z3. Since it is a ring homomorphism, we know to where [1] 2 Z9 must be sent. But this tells us where [2] = [1] + [1] must be sent (since it is a ring homomorphism), and [3], etc. How many ring homomor- phisms Z9 ! Z3 ⊕ Z3 exist? Are any of them onto (remember, there are nine elements in Z3 ⊕ Z3? Exercise. Do the same as above, but with Z6 and Z2 ⊕ Z3. Kernels. The kernel of a ring homomorphism φ : R ! S is the set defn fr 2 R φ(r) = 0g = ker φ. Examples: for evaluation φn : Z[x] ! Z: ker(φn) = f(x − n)g(x) g(x) 2 Z[x]g for `reduction mod n,' : Z ! Zn: ker = fnd d 2 Zg 2 for `projection to a coordinate' p1 : R ! R: ker p1 = f(r1; r2) r1 = 0g Proposition 2. A ring homomorphism φ : R ! S is 1-1 () ker φ = f0g. Proof. Suppose φ is 1-1 and let x 2 ker φ (x could be anything in ker φ). Then φ(0) = 0 = φ(x). Since φ is 1-1 this forces 0 = x. So anything that is in ker φ must be 0, so ker φ = f0g. Suppose that ker φ = f0g and let x; y 2 R be such that φ(x) = φ(y). We need to show that x must equal y. But φ(x) = φ(y) implies φ(x − y) = φ(x) − φ(y) = 0, so x − y 2 ker φ, and so x − y = 0. NOTES ON RINGS, MATH 369.101 3 Each of the kernels in examples above is a set of all multiples of some element. But not all ideals can be described as the set of multiples of one element: Define ζ : Z[x] ! Z2 by composing φ0 (evaluation at x = 0) with re- duction mod 2. Then ζ(f(x)) = 0 if and only if f(x) has an even constant coefficient. This is true for 2 2 Z[x] and for x 2 Z[x]. But it cannot be that ker ζ is the set of multiples (in Z[x]) of some f0(x) 2 Z[x]. Looking at 2, this f0 would need to be constant. It cannot be ±1, because then ker ζ would be all of Z[x] since every polynomial is something times ±1. So this would force f0(x) to be ±2, but then x cannot be a multiple of it (using only polynomials in Z[x]). A nice proposition. Proposition 3. If F is a field and φ : F ! R is a ring homomorphism (where 0 6= 1 in R), then φ must be 1-1. Proof. We know that φ(1) = 1R, the `1' in R. Choose x 2 ker φ. −1 Now if x 6= 0 then x exists in F. But then 1 = φ(1) = φ(xx−1) = φ(x)φ(x−1) = 0 · φ(x−1) = 0; which is untrue. So it must be that x = 0. Thus ker φ = f0g and so φ is 1-1 by Proposition 2. Ideals. It will be convenient to introduce the notion of a `ring without 1,' which is a set with addition and multiplication that would be a ring, except it does not have a multiplicative identity (a '1'). A simple example to think of is the set nZ of all integer multiples of some n. So for example, 2Z: this is closed under adding, multiplying, has zero in it, has additive inverses, etc. Let R be a commutative ring. An ideal I ⊂ R is a subset that is either a ring or a `ring without 1' and has the super-closed under multiplication property: for any r 2 R; and any x 2 I; rx 2 I: So an ideal always has a commutative ring R in which it sits. And if you multiply anything in R by something in I, you get something in I. Think again of 2Z. You can take any integer, multiply it by an even integer, and you get an even integer. Note: If an ideal I ⊂ R contains 1, then I = R: if 1 2 I, then 8r 2 R; r = r · 1 2 I. Also, any ideal contains 0. We call an ideal proper if I contains more than 0 and does not contain 1. Proposition 4. ker φ ⊂ R is an ideal for any ring h'sm φ : R ! S. Try checking this. You need to check ker φ satisfies all the ring axioms (except existence of 1) and the super-closed condition. The reason it will 4 NOTES ON RINGS, MATH 369.101 work is that multiplying by 0 gives 0 (and adding 0 to 0 gives 0), so you remain in the set of things sent by φ to 0. Alternatively, you could use this shorter way to check something is an ideal. Proposition 5. Let I ⊂ R be any subset, which is not empty. If the follow- ing conditions are satisfied, then I is an ideal: (1) if a; b 2 I then a − b 2 I. (2) if a 2 I and r 2 R then ra 2 I. So we have two ways to check if something is an ideal: • identify the set is the kernel of some homomorphism; • use Proposition 5. Proposition 6. Any ideal in Z and any ideal in F[x] is equal to the set of multiples of one element. Proof. For Z this statement is the content of Theorem 1.1.4. For F[x] this is implied by Theorem 4.2.2. For a little more detail: Start with all multiples of some given polynomial in the ideal. If this doesn't give the whole ideal, some poly. in the ideal is not a multiple of that one. These two have a gcd, and every combination of them is a multiple of that gcd (by Theorem 4.2.2). Eventually you stop getting a new gcd; for each new choice of polynomial either the degree of the gcd drops and you use induction, or the new polynomial has same gcd and has already been accounted for. Factor rings. A very generalized form of congruence classes. Fix an ideal I ⊂ R, where R a commutative ring. For r 2 R, write r + I to mean the set r + I = fr + x x 2 Ig: This set r + I is called a coset. You can choose different r's and get the same coset. For example, take the ideal 3Z ⊂ Z. So here I = 3Z.